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3 years ago

Mr Walden wrote the expression p^5 /q^0. He asked his students to write an equivalent expression of p^-5/q^0 in simplified form

Mathematics

2 answers:

kirill [66]3 years ago

7 0

Answer:

Anything to the power of zero equals one. Whatever q is will just cancel out because it's times one, so p^5

Step-by-step explanation:

Also is it p^5 or p^-5? I don't know if the answer would change but if the exponent is negative you could also say 1/p^5 I dont know if that's simplified form though

avanturin [10]3 years ago

7 0

Answer:

Brianna

Step-by-step explanation:

Hope this helps E2020

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Mike wrote down a number on a piece of paper and hid it from his friends. He rounded his number to different places and his frie

Naddik [55]

Answer: 145,998, 7,000, and 25,991

My guess :))

Step-by-step explanation:

145,998 7,002 4,506 26,991 12,497

<u>First, Mike rounded his number to the nearest ten</u>

146,000 7,000 4,510 27,000 12,500

<u>Then, he rounded his number to the nearest hundred.</u>

146,000 7,000 4,500 27,000 12,500

<u>Last, he rounded his number to the nearest thousand.</u>

146,000 7,000 5,000 27,000 12,000

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2 years ago

Bill found a pair of running shoes for $57. The regular price was $68.99. He also found a sweat suit that was originally $37.80.

mestny [16]

Answer: i believe bill saved $21.44

Step-by-step explanation: 68.99-57=11.99

37.80-25%=28.35

37.80-28.35=9.45

9.45+11.99=21.44

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3 years ago

Determine if the two figures are similar by using transformations.

katrin [286]

The answer is “translation followed by a dilation” :)

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3 years ago

In basketball, hang time is the time that both of your feet are off the ground during a jump. The equation for hang time is t =

const2013 [10]

Considering the hang time equation, it is found that Player 1 jumped 0.68 feet higher than Player 2.

<h3>What is the hang time equation?</h3>

The hang-time of the ball for a player of jump h is given by:

$t = 2\left(\frac{2h}{32}\right)^{\frac{1}{2}}$

The expression can be simplified as:

$t = 2\sqrt{\frac{h}{16}}$

For a player that has a hang time of 0.9s, the jump is found as follows:

$0.9 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.9}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.9}{2}\right)^2$

$h = 16\left(\frac{0.9}{2}\right)^2$

h = 3.24 feet.

For a player that has a hang time of 0.8s, the jump is found as follows:

$0.8 = 2\sqrt{\frac{h}{16}}$

$\sqrt{\frac{h}{16}} = \frac{0.8}{2}$

$(\sqrt{\frac{h}{16}})^2 = \left(\frac{0.8}{2}\right)^2$

$h = 16\left(\frac{0.8}{2}\right)^2$

h = 2.56 feet.

The difference is given by:

3.24 - 2.56 = 0.68 feet.

More can be learned about equations at brainly.com/question/25537936

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2 years ago

the temperature at noon was -27 degrees . the temperature increased by 13 degrees by 2pm . what was the temperature at 2pm

Aleks [24]

First, you would create an equation for this problem.
-27 + 13

Now, all you would have to do is solve it.
-27 + 13 = -14

By 2 pm the temperature was -14 degrees.

I hope this helps!

5 0

3 years ago

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